Tenth Grade Mathematics: Sample Spaces, Elementary Events, and Compound Events

The mathematical analysis of chance requires a rigorous set-theoretic language to classify every possible occurrence of a random phenomenon. A random experiment is any process whose exact individual outcome cannot be predicted with certainty in advance, but whose set of all possible outcomes is fully known. In this chapter, we formalize the concepts of the sample space, distinguish between elementary and compound events, and explore the fundamental addition property governing the probabilities of all elementary outcomes.

The Random Experiment and the Sample Space

An experiment is called a random experiment if it satisfies two conditions:
1. It has more than one possible outcome.

2. It is impossible to predict the exact outcome in advance of performing the trial.

Formal Definition of Sample Space:
The set of all possible outcomes of a random experiment is called the Sample Space
of the experiment, customarily denoted by the capital letter S.

Each individual element within the sample space is called a sample point or an elementary outcome. The total number of outcomes in the sample space is denoted by n(S).

Examples of Standard Sample Spaces:
1. Tossing a Single Coin:
   S = {H, T}                             ===> n(S) = 2

2. Tossing Two Distinguishable Coins:
   S = {HH, HT, TH, TT}                   ===> n(S) = 2^2 = 4

3. Rolling a Standard Six-Sided Die:
   S = {1, 2, 3, 4, 5, 6}                 ===> n(S) = 6

4. Rolling Two Standard Dice:
   S = {(1,1), (1,2), ..., (6,6)}         ===> n(S) = 6^2 = 36

Definition of an Event

In the formal language of set theory, any event is simply a subset of the sample space:

Formal Definition of an Event:
An Event E is any subset of the sample space S:
                            E subset S
We say that event E has occurred on a particular trial if the actual observed
outcome belongs to the set E.

The Empty and Total Events:
1. Impossible Event: The empty set phi is a subset of S.
   phi subset S  ===>  P(phi) = 0
2. Sure Event: The entire sample space S is a subset of itself.
   S subset S    ===>  P(S) = 1

Elementary Events vs. Compound Events

Events are structurally classified based on the number of sample points they contain:

1. Elementary (Simple) Event:
   An event having EXACTLY ONE single sample point:
   E = {omega_i}   where omega_i in S.
   Example: Rolling a standard die; getting a 4: E = {4}.

2. Compound Event:
   An event that contains TWO OR MORE distinct sample points:
   E = {omega_1, omega_2, ..., omega_k}   (k >= 2).
   A compound event can be decomposed into the union of elementary events.
   Example: Rolling a die; getting an even number: E = {2, 4, 6}.
            Here, E = {2} U {4} U {6}.

The Master Sum Property of Elementary Probabilities

A foundational theorem in probability theory establishes that when an experiment is partitioned into all its mutually exclusive elementary outcomes, their probabilities sum to unity:

Theorem (Sum of All Elementary Probabilities):
The sum of the probabilities of all the elementary events of an experiment
is strictly equal to 1:
            P(E1) + P(E2) + P(E3) + ... + P(Ek) = 1

Proof for Equally Likely Outcomes:
Let sample space S contain n equally likely outcomes: S = {e1, e2, ..., en}.
The probability of each elementary event is: P(ei) = 1 / n.
Sum of all elementary probabilities:
  P(e1) + P(e2) + ... + P(en) = (1/n) + (1/n) + ... + (1/n)  (n terms)
                               = n * (1/n) = 1.

Calculating Probabilities of Compound Events

If all elementary outcomes in a finite sample space S are equally likely, the probability of any compound event E is the sum of the probabilities of the elementary outcomes that comprise E:

Probability Formula:
             n(E)      Number of sample points favorable to E
     P(E) = ------ = -----------------------------------------
             n(S)      Total number of sample points in S

Chapter Exercises

Exercise 1: An experiment consists of tossing three coins simultaneously. (a) Write down the complete sample space S. (b) Find the number of total outcomes n(S). (c) Define the compound event of getting at least two heads and calculate its probability.

Exercise 2: A bag contains a red marble, a blue marble, and a yellow marble, all marbles being of the same size. Kritika takes out a marble from the bag without looking into it. State the elementary events and verify that the sum of their probabilities is 1.

Exercise 3: Two dice are thrown at the same time. Write down the sample space size and find the probability that the sum of the two numbers appearing on the top of the dice is: (a) 8, (b) 13, and (c) less than or equal to 12.

Solutions and Explanations

Solution 1: When tossing three coins, each coin has 2 possible outcomes (H or T). For part (a), the complete sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. For part (b), total number of outcomes is n(S) = 2^3 = 8. For part (c), let E be the event of getting "at least two heads" (meaning 2 heads or 3 heads). The favorable outcomes are E = {HHT, HTH, THH, HHH}. The number of favorable outcomes is n(E) = 4. The probability is P(E) = n(E) / n(S) = 4 / 8 = 1 / 2.

Solution 2: Let R denote the event of drawing a red marble, B denote drawing a blue marble, and Y denote drawing a yellow marble. The sample space is S = {R, B, Y}, with n(S) = 3. Since the marbles are of identical size, each outcome is equally likely. The three elementary events are: E1 = {R}, with P(E1) = 1/3; E2 = {B}, with P(E2) = 1/3; E3 = {Y}, with P(E3) = 1/3. Now sum the probabilities of all elementary events: P(E1) + P(E2) + P(E3) = 1/3 + 1/3 + 1/3 = 3/3 = 1. This verifies the master sum property.

Solution 3: When two dice are thrown, each die has 6 faces, so total outcomes n(S) = 6 * 6 = 36. For part (a), let E1 be the event that the sum is 8. The favorable pairs are (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2). The number of favorable outcomes is n(E1) = 5. P(E1) = 5 / 36. For part (b), let E2 be the event that the sum is 13. The maximum possible sum from two dice is 6 + 6 = 12. Since no pair can sum to 13, E2 is an impossible event: n(E2) = 0. P(E2) = 0 / 36 = 0. For part (c), let E3 be the event that the sum is less than or equal to 12. Because the maximum sum is 12, every single one of the 36 possible outcomes yields a sum less than or equal to 12. Thus, E3 is a sure event: n(E3) = 36. P(E3) = 36 / 36 = 1.